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In all the proofs of the determinant being the product of the eigenvalues I can find, including the one listed below, we start with the assumption that if $\lambda_1, \lambda_2,...,\lambda_n$ are the eigenvalues for $A$ then $\text{det}(A-\lambda I)=(-1)^n\prod_{i=1}^n(\lambda_i-\lambda)$. I understand that the $\lambda_i$ are roots of the char polynomial and thus each $(\lambda-\lambda_i)$ will be a factor, but how do we know there are no other factors? Specifically, how do we know there are no other factors which are polynomials with no roots?

Show that the determinant of $A$ is equal to the product of its eigenvalues

mdcq
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tmpys
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1 Answers1

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I suppose that your matrix $A$ is a $n\times n$ matrix and that $\lambda_1,\ldots,\lambda_n$ are all the eigenvalues, eventually with repetitions (for instance, if $A=\operatorname{Id}$, then each $\lambda_k$ is equal to $1$). If that is so, then $\deg\det(A-\lambda\operatorname{Id})=n$ and therefore there is no room for other factors. Besides, this proof assumes that we are working in an algebraically clased field, such as $\mathbb C$, and in such a field there are no non-constant polynomials without roots.

José Carlos Santos
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